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24 Aug 2022, 02:06
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D
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Difficulty:
75%
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Question Stats:
57% (03:19) correct
43%
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wrong
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In the diagram above, triangle ABC has a right angle at B and a perimeter of 120. Line segment BD is perpendicular to AC and has a length of 24. AB > BC. What is the ratio of the area of triangle ABD to the area of triangle BDC?
A. 9/7
B. 16/13
C. 16/11
D. 16/9
E. 16/7
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Attachment:
Triangle.png
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24 Aug 2022, 04:01
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Was a bit tricky for me, tried to assume what the triangle could be and was able to find possible answer.
Since it is a right angled triangle assuming 3,4,5 as triplets of the triangle we find that a perimeter of 120 can be obtained by multiplying each side by 10.
So sides of triangle become AB =40; BC=30 (because AB>BC); and AC=50
Triangle ABD is similar to triangle BDC because BD is perpendicular to AC. We get,
AD*CD = BD^2 = 24^2
CD = AC-AD = 50-AD
AD*(50-AD) = 24^2
Let, AD=x
x^2 - 50x + 24^2 = 0
x^2 - 32x - 18x + 576 =0
(x-32)(x-18) = 0
=> AD= 32 or AD=18
Because AB>BC we have AD>CD
=> AD=32 and CD=18
A(△ABD) / A(△BDC) = AD*BD / CD*BD = AD/CD = 32/18 = 16/9
Answer: D
Since it is a right angled triangle assuming 3,4,5 as triplets of the triangle we find that a perimeter of 120 can be obtained by multiplying each side by 10.
So sides of triangle become AB =40; BC=30 (because AB>BC); and AC=50
Triangle ABD is similar to triangle BDC because BD is perpendicular to AC. We get,
AD*CD = BD^2 = 24^2
CD = AC-AD = 50-AD
AD*(50-AD) = 24^2
Let, AD=x
x^2 - 50x + 24^2 = 0
x^2 - 32x - 18x + 576 =0
(x-32)(x-18) = 0
=> AD= 32 or AD=18
Because AB>BC we have AD>CD
=> AD=32 and CD=18
A(△ABD) / A(△BDC) = AD*BD / CD*BD = AD/CD = 32/18 = 16/9
Answer: D
wadhwakaran
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Schools: INSEAD Aug'24 intake Oxford Said'25 Judge '25 Imperial'25 HEC Sep'24 Intake LBS '26 ISB '27
GMAT 1: 700 Q50 V35
Posts: 260
24 Aug 2022, 22:05
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Since, the perimeter of the triangle ABC is 120
AB=x, BC=y, AC=z and AB>BC (x>y)
BD=24
x+y+z=120
Pythagorean Triplet (3,4,5 has a sum of 12) meaning thereby that we can replace x,y,z by a multiple of this triplet.
x=40, y=30 and z=50
Now that Tr (ABD) is similar to Tr (BCD)
Ratio of Area of similar triangles = ratio of square of corresponding sides
Ar (ABD)/Ar (BCD)=AB*AB/BC*BC=40*40/30*30=16/9
Therefore option D is the answer.
AB=x, BC=y, AC=z and AB>BC (x>y)
BD=24
x+y+z=120
Pythagorean Triplet (3,4,5 has a sum of 12) meaning thereby that we can replace x,y,z by a multiple of this triplet.
x=40, y=30 and z=50
Now that Tr (ABD) is similar to Tr (BCD)
Ratio of Area of similar triangles = ratio of square of corresponding sides
Ar (ABD)/Ar (BCD)=AB*AB/BC*BC=40*40/30*30=16/9
Therefore option D is the answer.
